Question

\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]

Answer 1

\[\int\left( \frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} \right) dx\]
\[\text{Let x + 1 }= t\]
\[ \Rightarrow x = t - 1\]
\[ \Rightarrow 1 = \frac{dt}{dx}\]
\[ \Rightarrow dx = dt\]
\[Now, \int\left( \frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} \right) dx\]
\[ = \int\left[ \frac{\left( t - 1 \right)^2 + 3\left( t - 1 \right) + 1}{t^2} \right]dt\]
\[ = \int\left( \frac{t^2 - 2t + 1 + 3t - 3 + 1}{t^2} \right)dt\]
\[ = \int\left( \frac{t^2 + t - 1}{t^2} \right)dt\]
\[ = \int\left( 1 + \frac{1}{t} - t^{- 2} \right) dt\]
\[ = t + \text{ log }\left| t \right| - \frac{t^{- 2 + 1}}{- 2 + 1} + C\]
\[ = t + \text{ log }\left| t \right| + \frac{1}{t} + C\]
\[ = x + 1 + \text{ log     }\left| x + 1 \right| + \frac{1}{x + 1} + C\]
\[\text{ Let 1 + C  }= C'\]
\[ = x + \text{ log }\left| x + 1 \right| + \frac{1}{x + 1} + C'\]